Question

Find the sum to n terms of the series whose nth term is
(i)n(n+1)(n+3)
(ii)n²-2n+2+2​

Answer 1

Answer

1. $\dfrac{1}{12}n(n+1)(n+2)(3n+13)$
2. $\dfrac{2n^3-3n^2+7n-12}{6} +2^{n+2}$

Definition of Sigma

The sum of the first n terms of a series $a_n$ is $\displaystyle\sum^{n}_{k=1}a_k$.

The sum inside Sigma can be decomposed.

The constant product inside Sigma can be pulled out.

Calculation

The calculation of (1) is as follows.

$\displaystyle\sum^{n}_{k=1}k(k+1)(k+3)$

$=\displaystyle\sum^{n}_{k=1}(k^3+4k^2+3k)$

$=\displaystyle\sum^{n}_{k=1}k^3+\displaystyle\sum^{n}_{k=1}4k^2+\displaystyle\sum^{n}_{k=1}3k$

$=\{\dfrac{n(n+1)}{2} \}^2+4\{\dfrac{n(n+1)(2n+1)}{6} \}+3\{\dfrac{n(n+1)}{2} \}$

$=\dfrac{n(n+1)}{2} \{\dfrac{n(n+1)}{2} +\dfrac{4(2n+1)}{3} +\dfrac{3}{2} \}$

$=\dfrac{n(n+1)}{2} (\dfrac{n^2}{2} + \dfrac{19n}{6} + \dfrac{17}{6})$

$=\dfrac{n(n+1)(3n^2 + 19n + 17)}{12}$

$=\dfrac{1}{12}n(n+1)(n+2)(3n+13)$

The calculation of (2) is as follows.

$\displaystyle\sum^{n}_{k=1}(k^2-2k+2)+\displaystyle\sum^{n}_{k=1}2^k$

$=\displaystyle\sum^{n}_{k=1}(k-1)^2+\displaystyle\sum^{n}_{k=1}1+\displaystyle\sum^{n}_{k=1}2^k$

$=\dfrac{(n-1)n(2n-1)}{6} +n+\dfrac{2(2^{n+1}-1)}{2-1}$

$=\dfrac{(n-1)n(2n-1)}{6} +\dfrac{6n}{6} +2^{n+2}-2$

$=\dfrac{n(2n^2-3n+1+6)-12}{6} +2^{n+2}$

$=\dfrac{2n^3-3n^2+7n-12}{6} +2^{n+2}$

More information

You can use identities to calculate sigma.

For example, completing the square is used in (2).

We can solve (1) with $(k+1)^3=k^3+3k^2+3k+1$.

Like,

$\displaystyle\sum^{n}_{k=1}(k^3+4k^2+3k)$

$=\displaystyle\sum^{n}_{k=1}(k^3+3k^2+3k+1)+\displaystyle\sum^{n}_{k=1}(k^2-1)$

$=\displaystyle\sum^{n}_{k=1}(k+1)^3+\displaystyle\sum^{n}_{k=1}k^2+\displaystyle\sum^{n}_{k=1}-1$ and so on.